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Everything else is correct except equations (2.1) to (3.2). Equation (4) cannot be used to since the Hamiltonian is time dependent. What you have to do is take the derivative of (6.1) and substitute equation (6.2). By doing this you will get $$\frac{d^2X(t)}{dt^2}+\omega^2X(t)=\frac{qE_0}{m}\cos{\omega't}$$ I have solved this equation by assuming $\omega=\omega'=q=E_0=m=1$ and $X(0)=1,\dot{X}(0)=0$ for simplicity and got the result as $$X(t)=\frac{1}{4}(2\cos^3t+2\cos t+2t\sin t+\sin{2t}\sin{t})$$ On simplifying one gets $$X(t)=\frac{1}{2}t\sin{t}+\cos{t}$$

Hope this helps.

##Edit The answer given by @CosmasZachos is more accurate and general since I have taken $\omega=\omega'$. Though this solution is correct (checked it on Mathematica) it works only when the condition is satisfied, while his solution works in general.

Everything else is correct except equations (2.1) to (3.2). Equation (4) cannot be used to since the Hamiltonian is time dependent. What you have to do is take the derivative of (6.1) and substitute equation (6.2). By doing this you will get $$\frac{d^2X(t)}{dt^2}+\omega^2X(t)=\frac{qE_0}{m}\cos{\omega't}$$ I have solved this equation by assuming $\omega=\omega'=q=E_0=m=1$ and $X(0)=1,\dot{X}(0)=0$ for simplicity and got the result as $$X(t)=\frac{1}{4}(2\cos^3t+2\cos t+2t\sin t+\sin{2t}\sin{t})$$ On simplifying one gets $$X(t)=\frac{1}{2}t\sin{t}+\cos{t}$$

Hope this helps.

Edit

The answer given by @CosmasZachos is more accurate and general since I have taken $\omega=\omega'$. Though this solution is correct (checked it on Mathematica) it works only when the condition is satisfied, while his solution works in general.

Everything else is correct except equations (2.1) to (3.2). Equation (4) cannot be used to since the Hamiltonian is time dependent. What you have to do is take the derivative of (6.1) and substitute equation (6.2). By doing this you will get $$\frac{d^2X(t)}{dt^2}+\omega^2X(t)=\frac{qE_0}{m}\cos{\omega't}$$ I have solved this equation by assuming $\omega=\omega'=q=E_0=m=1$ and $X(0)=1,\dot{X}(0)=0$ for simplicity and got the result as $$X(t)=\frac{1}{4}(2\cos^3t+2\cos t+2t\sin t+\sin{2t}\sin{t})$$ On simplifying one gets $$X(t)=\frac{1}{2}t\sin{t}+\cos{t}$$

Hope this helps.

Everything else is correct except equations (2.1) to (3.2). Equation (4) cannot be used to since the Hamiltonian is time dependent. What you have to do is take the derivative of (6.1) and substitute equation (6.2). By doing this you will get $$\frac{d^2X(t)}{dt^2}+\omega^2X(t)=\frac{qE_0}{m}\cos{\omega't}$$ I have solved this equation by assuming $\omega=\omega'=q=E_0=m=1$ and $X(0)=1,\dot{X}(0)=0$ for simplicity and got the result as $$X(t)=\frac{1}{4}(2\cos^3t+2\cos t+2t\sin t+\sin{2t}\sin{t})$$ On simplifying one gets $$X(t)=\frac{1}{2}t\sin{t}+\cos{t}$$

Hope this helps.

##Edit The answer given by @CosmasZachos is more accurate and general since I have taken $\omega=\omega'$. Though this solution is correct (checked it on Mathematica) it works only when the condition is satisfied, while his solution works in general.

Everything else is correct except equations (2.1) to (3.2). Equation (4) cannot be used to since the Hamiltonian is time dependent. What you have to do is take the derivative of (6.1) and substitute equation (6.2). By doing this you will get $$\frac{d^2X(t)}{dt^2}+\omega^2X(t)=\frac{qE_0}{m}\cos{\omega't}$$ I have solved this equation by assuming $\omega=\omega'=q=E_0=m=1$ and $X(0)=1,\dot{X}(0)=0$ for simplicity and got the result as $$X(t)=\frac{1}{4}(2\cos^3t+2\cos t+2t\sin t+\sin{2t}\sin{t})$$

Hope this helps.

Everything else is correct except equations (2.1) to (3.2). Equation (4) cannot be used to since the Hamiltonian is time dependent. What you have to do is take the derivative of (6.1) and substitute equation (6.2). By doing this you will get $$\frac{d^2X(t)}{dt^2}+\omega^2X(t)=\frac{qE_0}{m}\cos{\omega't}$$ I have solved this equation by assuming $\omega=\omega'=q=E_0=m=1$ and $X(0)=1,\dot{X}(0)=0$ for simplicity and got the result as $$X(t)=\frac{1}{4}(2\cos^3t+2\cos t+2t\sin t+\sin{2t}\sin{t})$$ On simplifying one gets $$X(t)=\frac{1}{2}t\sin{t}+\cos{t}$$

Hope this helps.